For all $n\geq 1$, prove with mathematical induction
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$
So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to go about the simplification.. right now I have
LHS: $2-\frac{1}{k}+\frac{1}{(k+1)^2} \leq 2-\frac{1}{k+1}$
Should I try to find common denominators for the left? Step by step explanation please!
you simply have $$2-\frac{1}{k}+\frac{1}{(k+1)^2} \\ =2-\{\frac{1}{k}-\frac{1}{(k+1)^2}\} \\ =2-\{\frac{(k+1)^2-k}{k(k+1)^2}\}\\ =2-\{\frac{k^2+k+1}{k(k+1)^2}\}\\ \leq 2-\frac{k(k+1)}{k(k+1)^2}$$
Since $$k^2+k+1 \gt k^2+k\\ \frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\ -\frac{k^2+k+1}{k(k+1)^2} \le -\frac{k^2+k}{k(k+1)^2}$$
This exercise shows that the sum of the reciprocals of the squares converges to something at most $2$; in fact, the series converges to $\frac{\pi^2}{6}$.
For $n\geq 1$, denote the statement in the exercise by $$ S(n) : 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} \leq 2 - \frac{1}{n}. $$
Base step ($n=1$): Since $1=2-\frac{1}{1}, S(1)$ holds.
Induction step: Fix some $k\geq 1$ and suppose that $S(k)$ is true. It remains to show that $$ S(k+1) : 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{k^2} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1} $$ holds. Starting with the left side of $S(k+1)$, \begin{align} 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2} &\leq 2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad(\text{by } S(k))\\[1em] &= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em] &= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em] &\leq 2-\frac{1}{k+1},\quad(\text{since } k\geq 1, k^2-k\geq 0) \end{align} the right side of $S(k+1)$. Thus $S(k+1)$ is true, thereby completing the inductive step.
By mathematical induction, for any $n\geq 1$, the statement $S(n)$ is true.
